【发布时间】:2018-05-23 02:40:11
【问题描述】:
我正在尝试在 Swift 3 中加密一个字符串,而我的加密每次都给出不同的输出。这是为什么? (我在python中尝试过类似的加密,加密后的输出总是一样的)。
这是我的 Swift 3 aesEncrypt 函数:
func aesEncrypt(key:String, iv:Array<Any>, options:Int = kCCOptionPKCS7Padding) -> String? {
if let keyData = sha256(string:key),
let data = self.data(using: String.Encoding.utf8),
let cryptData = NSMutableData(length: Int((data.count)) + kCCBlockSizeAES128) {
let keyLength = size_t(kCCKeySizeAES128)
let operation: CCOperation = UInt32(kCCEncrypt)
let algorithm: CCAlgorithm = UInt32(kCCAlgorithmAES128)
let options: CCOptions = UInt32(options)
var numBytesEncrypted :size_t = 0
let cryptStatus = CCCrypt(operation,
algorithm,
options,
(keyData as NSData).bytes, keyLength,
iv,
(data as NSData).bytes, data.count,
cryptData.mutableBytes, cryptData.length,
&numBytesEncrypted)
// ADDED PRINT STATEMENTS
print("keyData")
print(keyData)
print("\(keyData as NSData)")
print("iv")
print(iv)
var hex_iv = toHexString(arr: iv as! [UInt8])
print(hex_iv)
print("data")
print(data)
print("\(data as NSData)")
print("encryption: cryptdata")
print(cryptData)
print("encryption: num bytes encrypted")
print(numBytesEncrypted)
if UInt32(cryptStatus) == UInt32(kCCSuccess) {
cryptData.length = Int(numBytesEncrypted)
let base64cryptString = cryptData.base64EncodedString(options: .lineLength64Characters)
return base64cryptString
}
else {
return nil
}
}
return nil
}
当我尝试使用 initial_string = "hello" 运行以下代码时,我每次都会得到不同的加密输出字符串。
let iv [UInt8](repeating: 0, count: 16)
let key = "sample_key"
let initial_string = "hello"
let encryptedString = initial_string.aesEncrypt(key: key, iv: iv)
print("Encrypted string")
print(encryptedString)
第一次运行带有“hello”字符串的代码的示例输出:
keyData
32 bytes
<d5a78c66 e9b3ed40 b3a92480 c732527f 1a919fdc f68957d2 b7e9218f 6221085d>
iv
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
00-00-00-00-00-00-00-00-00-00-00-00-00-00-00-00
data
5 bytes
<68656c6c 6f>
encryption: cryptdata
<b17d67fc 26e3f316 6a2bdfbf 9d387c2d 00000000 00>
encryption: num bytes encrypted
16
Encrypted string
Optional("sX1n/Cbj8xZqK9+/nTh8LQ==")
第二次使用“hello”字符串运行代码的示例输出:
keyData
32 bytes
<d5a78c66 e9b3ed40 b3a92480 c732527f 1a919fdc f68957d2 b7e9218f 6221085d>
iv
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
00-00-00-00-00-00-00-00-00-00-00-00-00-00-00-00
data
5 bytes
<68656c6c 6f>
encryption: cryptdata
<01b9f69b 45deb31d eda46c2d dc9ad9e8 00000000 00>
encryption: num bytes encrypted
16
Encrypted string
Optional("Abn2m0Xesx3tpGwt3JrZ6A==")
你能告诉我为什么每次对于同一个键、iv 和字符串的输出都不同吗?谢谢!
【问题讨论】:
-
看看stackoverflow.com/a/37681510/1187415中的代码,应该可以的。
-
该代码是扩展但缺少扩展代码。但是在回答您的问题时:因为每次输入都不同。在调用之前打印
keyData、data和iv,在调用CCCrypt之后打印cryptData-- 十六进制。将其添加到问题中,但我猜你会在那里找到问题。 -
@zaph 我刚刚添加了打印语句,如果有帮助请告诉我!
标签: python swift encryption swift3 aes